Let $Q$ be a special orthogonal matrix. Show that $Q(u\times v)=Q(u)\times Q(v)$ for any vectors $u, v\in\mathbb R^3$. 
Let $Q$ be a $3\times3$ special orthogonal matrix. Show that $Q(u\times v)=Q(u)\times Q(v)$ for any vectors $u, v\in\mathbb R^3$.

I have no idea how to start. I'm not sure if $Q(u)\cdot Q(V)=Q(u\cdot v)$ would helps. Please give me some help. Thanks.
 A: The cross-product $u\times v$ is the unique vector such that
$$
\det(u,v,w)=(u\times v)\cdot w\qquad \forall w
$$
where $\det(u,v,w)$ is the determinant of the $3\times 3$ matrix whose columns are $u,v,w$ in this order, that  is the determinant of the linear map that sends the canonical basis to $(u,v,w)$. That's a common definition of the cross-product. See below if needed.
Recall that $Q$ special orthogonal means $Q^T=Q^{-1}$ and $\det Q=1$.

We need to prove that $Q^T(Qu\times Qv)=u\times v$. So let us compute
  $$
Q^T(Qu\times Qv)\cdot w=(Qu\times Qv)\cdot Qw=\det(Qu,Qv,Qw)=\det Q\det(u,v,w)=\det(u,v,w).
$$
  By the uniqueness defining $u\times v$, this proves $Q^T(Qu\times Qv)=u\times v$, i.e. $Qu\times Qv=Q(u\times v)$.

Note: the same argument shows more generally that, as mentioned by lhf and wikipedia,
$$
M^T(Mu\times Mv)=(\det M) u\times v\quad\Rightarrow \quad (Mu\times Mv)=(\det M) M^{-T}(u\times v)
$$
for every invertible $3\times 3$ matrix $M$, where $M^{-T}=(M^{-1})^T=(M^T)^{-1}$. The formula on the left is true for every matrix $M$ and is just $0=0$ in the singular case, since we have $Mu\times Mv=0$ for every $u,v$ in this case.

The fact that the identity $\det(u,v,w)=(u\times v)\cdot w$ is satisfied by every $u,v,w$ can be checked directly, by computations, from the determinant definition of $u\times v$. Another way to see it is to note that the map $(u,v,w)\longmapsto (u\times v)\cdot w$ is multilinear, anti-symmetric (or alternating), and sends the canonical basis to $1$, whatever definition of the cross-product you might have. So it must be the determinant. Uniqueness of $u\times v$ satisfying the identity follows from $(\mathbb{R}^{3})^\perp=\{0\}$, as $w_1\cdot w=w_2\cdot w$ for every $w$ implies $(w_1-w_2)\cdot w=0$ for every $w$, in particular for $w=w_1-w_2$, whence $\|w_1-w_2\|^2=0$.
A: Maybe it will be easiest to show this explicitly for the basis vectors $ \{ (1,0,0) \cdots \} $ , and then the general case follows from linearity of all things involved. It will be useful to note that if $ \vec{Q_1}, \vec{Q_2}, \vec{Q_3} $ are the column vectors of $ Q $, then the fact that $ \det(Q) = 1 = \vec{Q1} \cdot (\vec{Q_2} \times \vec{Q_3}) $ gives the "right hand rule" that $\vec{Q_1} \times \vec{Q_2} = \vec{Q_3} $. 
Please note, I have been very sloppy and did not check the signs and orders of things. You should check that all formulae are indeed right.
